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Chapter 16: Problem 25

By what factor does the rate change in each of the following cases (assuming constant temperature)? (a) A reaction is first order in reactant \(\mathrm{A}\), and \([\mathrm{A}]\) is doubled. (b) A reaction is second order in reactant \(\mathrm{B},\) and \([\mathrm{B}]\) is halved. (c) A reaction is second order in reactant \(\mathrm{C},\) and \([\mathrm{C}]\) is tripled.

Short Answer

Expert verified
a) 2, b) 1/4, c) 9.

Step by step solution

01

Understand Rate Laws

The rate law for a reaction tells us how the rate of reaction depends on the concentration of reactants. For a reaction that is first order in reactant \(\text{A}\), the rate law is \( \text{rate} = k[\text{A}] \) where k is the rate constant. For a reaction that is second order in reactant \(\text{B}\), the rate law is \( \text{rate} = k[\text{B}]^2 \). For a reaction that is second order in reactant \(\text{C}\), the rate law is \( \text{rate} = k[\text{C}]^2 \).
02

Analyze Case (a)

For a reaction that is first order in reactant \(\text{A}\), the rate law is \( \text{rate} = k[\text{A}] \). If the concentration of \(\text{A}\) is doubled, the new rate will be \( k(2[\text{A}]) = 2k[\text{A}] \). Thus, the rate doubles. The factor by which the rate changes is 2.
03

Analyze Case (b)

For a reaction that is second order in reactant \(\text{B}\), the rate law is \( \text{rate} = k[\text{B}]^2 \). If the concentration of \(\text{B}\) is halved, the new rate will be \( k\bigg( \frac{[\text{B}]}{2} \bigg)^2 = k\frac{[\text{B}]^2}{4} \). Thus, the rate is reduced by a factor of 4. The factor by which the rate changes is \(\frac{1}{4} \).
04

Analyze Case (c)

For a reaction that is second order in reactant \(\text{C}\), the rate law is \( \text{rate} = k[\text{C}]^2 \). If the concentration of \(\text{C}\) is tripled, the new rate will be \( k(3[\text{C}])^2 = k9[\text{C}]^2 \). Thus, the rate increases by a factor of 9. The factor by which the rate changes is 9.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Order Reaction
In first order reactions, the reaction rate is directly proportional to the concentration of one reactant. This means that if the concentration of reactant \(A\) changes, the reaction rate will change in the same way. The rate law for a first order reaction is given by \( \text{rate} = k[\text{A}] \). Here, \(k\) is the rate constant, and \[ \text{A} \] is the concentration of the reactant. For example, if you double the concentration of \[ \text{A} \], the rate of the reaction will also double. Similarly, halving the concentration of \[ \text{A} \] will halve the reaction rate.

Key Takeaways:
  • The rate of a first order reaction is proportional to the reactant concentration.
  • The rate constant \(k\) remains unaffected by changes in concentration.
  • First order reactions have a characteristic exponential decay of reactant concentration over time.
Second Order Reaction
Second order reactions are characterized by a reaction rate that is proportional to the square of the concentration of one reactant or to the product of the concentrations of two different reactants. The general rate law for a second order reaction is \( \text{rate} = k[\text{B}]^2 \) for a single reactant or \( \text{rate} = k[\text{A}][\text{B}] \) when two different reactants are involved. In our example, consider reactant \[ \text{B} \]:
  • If \[ \text{B} \] is halved, the rate becomes \[\frac{ \text{rate} = k\bigg(\frac{[\text{B}]}{2}\bigg)^2}{ \text{rate} = k\frac{[\text{B}]^2}{4} \].
  • If \[ \text{B} \] is doubled, the rate becomes \[k(2[\text{B}])^2 = k4[\text{B}]^2 \].

Key Points:
  • The rate of second order reactions is more sensitive to changes in reactant concentration compared to first order reactions.
  • If the concentration of the reactant is changed, the effect on the rate is squared.
  • Second order reactions often have different units for the rate constant \(\text{k}\) compared to first order reactions.
Rate Constant
The rate constant, denoted as \(k\), is a crucial component in determining the speed of a reaction according to a rate law. For a given reaction, the rate constant is a proportionality factor that links the concentration of reactants to the reaction rate. It remains constant as long as the temperature is constant and does not change with reactant concentration changes.

Characteristics of the Rate Constant:
  • It is specific to a particular reaction at a given temperature.
  • The units of \(k\) vary depending on the reaction order. For first order reactions, the units are \( \text{time}^{-1} \), while for second order reactions, the units are \( \text{concentration}^{-1} \text{time}^{-1} \).
  • The value of the rate constant gives insight into the speed of the reaction – higher \(k\) values indicate faster reactions.
The rate constant can also change with temperature, following the Arrhenius equation \[ k = Ae^{-\frac{E_a}{RT}} \], where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

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Most popular questions from this chapter

The compound \(\mathrm{AX}_{2}\) decomposes according to the equation \(2 \mathrm{AX}_{2}(g) \rightarrow 2 \mathrm{AX}(g)+\mathrm{X}_{2}(g) .\) In one experiment, \(\left[\mathrm{AX}_{2}\right]\) was measured at various times and these data were obtained: $$ \begin{array}{cc} \text { Time (s) } & {\left[A X_{2}\right](\mathrm{mol} / \mathrm{L})} \\ \hline 0.0 & 0.0500 \\ 2.0 & 0.0448 \\ 6.0 & 0.0300 \\ 8.0 & 0.0249 \\ 10.0 & 0.0209 \\ 20.0 & 0.0088 \end{array} $$ (a) Find the average rate over the entire experiment. (b) Is the initial rate higher or lower than the rate in part (a)? Use graphical methods to estimate the initial rate.

The mathematics of the first-order rate law can be applied to any situation in which a quantity decreases by a constant fraction per unit of time (or unit of any other variable). (a) As light moves through a solution, its intensity decreases per unit distance traveled in the solution. Show that \(\ln \left(\frac{\text { intensity of light leaving the solution }}{\text { intensity of light entering the solution }}\right)\) \(=-\) fraction of light removed per unit of length \(\times\) distance traveled in solution (b) The value of your savings declines under conditions of constant inflation. Show that \(\ln \left(\frac{\text { value remaining }}{\text { initial value }}\right)\) \(=-\) fraction lost per unit of time \(\times\) savings time interval

Is it possible for more than one mechanism to be consistent with the rate law of a given reaction? Explain.

You are studying the reaction \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(g)\) to determine its rate law. Assuming that you have a valid experimental procedure for obtaining \(\left[\mathrm{A}_{2}\right]\) and \(\left[\mathrm{B}_{2}\right]\) at various times, explain how you determine (a) the initial rate, (b) the reaction orders, and (c) the rate constant.

At body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the rate constant of an enzyme-catalyzed decomposition is \(2.3 \times 10^{14}\) times that of the uncatalyzed reaction. If the frequency factor, \(A,\) is the same for both processes, by how much does the enzyme lower the \(E_{\mathrm{a}}\) ?
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